139016545-23-Ray-Optics

Ray Optics

* Light " Reflection of Light • Spherical Mirrors " Refraction * Total Internal Reflection .. Refraction from a Spherical Surface

• .. ,. "

Lens Refraction Through a Prism Dispersion Optical Instruments ~· Resolving Power of a Microscop~~ " Resolvi~g Power of Telescope

------------------,~.~------------------·

t> light The light is that form of energy which makes objects visible to our eyes. The branch of physics which deals with nature of light, its sources, properties, measurement, effects and vision is called "optics". For the sake of convenience, study of optics is generally divided into two parts namely (i) geometrical optics or ray optics, and (ii) wave optics. This chapter deals with the geometrical optics.

*> Reflection of Light When a beam of light is incident on a polished interface, it is thrown back in same medium. This phenomenon is called reflection. In reflection the frequency, speed and wavelength do not change, but a phase change may occur depending on the nature of reflecting surface. Experimentally it is found that the rays corresponding to the incident and reflected waves make equal angles with the normal to the surface . Thus, the two laws of reflection can be summarized as under. 1. Li = Lr

2. Incitlent ray, reflected ray and normal lie on the same plane.

886

Chapter 23 • Ray Optics Normal

1

J":

If

>bs0·t-,

.(\iil Note · The above two laws of reflection can b.e appli.ed to the reflecting surfaces · wh}ch a(e not ev.en horizontal.

Instance 2 Find velocity of image when object and mirror both are moving toward each other with velocity 2 ms-1 and 3 ms- 1 respectively. (b) --Bms-1 (a) 8ms-1 (c) -5ms-1 (d) 5ms- 1 Interpret

Here

9rh r(g,.

Vo-VM=-(Vl-VM)

1t811e·ction from Plane Mirror (Surface} In case of reflection from plane surface such as plane mirror (i) The image is always erect, virtual and of exactly the same size as the object. The image is formed as much behind the mirror as the object is in from of it. (ii) The image is laterally inverted. (iii) If keeping the incident ray fixed, the plane mirror is rotilted through an angle e, the reflected ray turns through double the angle ie, 28 in that very direction. (iv) If the object is fixed and the mirror moves relative to the · ,object with a speed v, the image moves with a speed 2v ~elative to the object. (v) If the mirror is fixed and the object moves relative to the mirror with a speed v, the image also moves with the same speed..y relative to the mirror. (vi) Deviation suffered by a light ray incident at an angle i is given by 8 = (180 - 2i) , (vii) If there are two mirrors inclined at an angle 8, the total number of images formed for an object kept between 2 the two is equal to 7t or ( ~7t - 1} which ever is odd. 8 (viii) The minimum size of a mirror required to see the full image of a person, is half the height of the person. (ix) If a plane mirror is rotated about an axis perpendicular to plane of mirror then reflected ray image do not rotate.

Instance 1 The minimum size of the mirror fixed on the wall of a room in which an observer at the centre of room may see the full image of the wall of heightJLbehind him is (a)

hx z-(2x+y)

Size of mirror, M M -

~

(b)

3

(c) 2h

~ 2

(d) h

3

lnterp.ret

From

~ 0' M1M2

M 1M 2 = - x -

h

2x +y

and O'AB

=> (+2ms-

1

1 ) - ( -3ms- )

=~vi+ (-3)

Instance 3 Two plane mirrors are inclined at 30° as shown in figure. A light ray is incident at angle 45°. Find total deviation produced by combination of mirror after two successive reflection.

(a) (c)

60° i 50° i

(b) 58° i (d) 68° i

Deviation at mirror M1 ,8 1 = 180°-2 x 45°= 90° i

Interpret

Deviation at mirror M2 ,8 2 = 180°-2 x 15° = 150°t Total deviation 8 = 82 -8 1 = 150°- 90° = 60° i

• Spherical Mirrors Mirrors having their reflecting surface spherical are called spherical mirrors. Spherical mirrors are of two types (i) Concave mirror If reflection takes place from the inner surface, the mirror is called concave [Fig. (a)]. (ii) Convex mirror If reflection takes place from the outer surface , the mirror is called convex [Fig. (b)] . A Incident light

Incident light

---+

-----...--....-=: P----.. +ve

h

---+

,

(a) Concave mirror

--=p~------

-

c

(b) Convex mirror

887

Chapter 23 • Ray Optics

Definitions of Some Terms Related to Spherical Mirrors

Ray Tracing

Centre and radius of curvature The centre of curvature and radius of curvature of a mirror are the centre and radius of the sphere of which the mirror is part. In the given figure, AC is the radius of curvature and C, the centre of curvature.

In geometrical optics, to locate the image of an object. Tracing of a ray as it reflects or refracts, is very important. 1. A ray going through centre of curvature is reflected back along the same direction.

a

Concave

Convex

~·

c

Pole Pole of the spherical mirror is mid point of its reflecting surface. In figure it is shown by P.

c

2. A ray parallel to principal axis is reflected through the focus, and vice-versa. Also, mutually parallel rays. !\{!]1t. reflection intersect on the focal plane.

Principal ,axis The principal axis of a spherical mirror is the line joining the pole and centre of curvature. In the figure PC is principal axis. Principal focus Principal focus is a point on the principal axis of the mirror at which the light rays coming parallel to principal axis actually meet after reflection or appear to meet.

F

.E)

3. The light corning through the focus of rriirror or coming towards focus, becomes parallel to principal axis. ;)

c F

(b) Convex mirror

(a) Concave mirror 0

For concave mirror focus is infront of the mirror, while for convex mirror focus is behind the mirror. Focus of concave mirror is real, while focus of convex mirror is virtual.

Sign Convention for Mirrors According to the sign convention (i) Origin should be placed at the pole (P). (ii) All distances should be measured from the pole (P). (iii) Object distance is denoted by u, image distance by v, focal length by f and radius of curvature by R. (iv) Distance measured in the direction of incident ray are taken as positive while in the direction opposite of incident ray are taken negative.

Focal length The distance between pole and focus of a spherical mirror is called its focal length. It is represented byf

f

ie,

Table 23.1

1.

= !!_

2

Image Formation by Concave Mirror

p

At infinity

Real inverted, very small [m < < -1], at F t-

2.

Between infinity and

p

C

M'

Real, inverted, diminished (m < -1) between F and C

888

Chapter 23 • Ray Optics

M

3.

Real, inverted, equal in size [m = -1 ] at C

AtC

M'

M

4.

Real, inverted and very large (m > -1) between 2F and iiliini.ty

Between F and C

M'

\ ~

5.

Atf

Real, inverted, very large [m infinity

6.

Between F and P

Virtual, erect, large in size (m > + 1) behind the mirror

Table 23.2

1.

( -oo)] at

Image Formation by Convex Mirror

Virtual, erect, very small (0 < m < < + 1) at F

At infinity

c

\

M'

)

2.

Virtual, erect, diminished (m < + 1) between PandF

In front of mirror

c M'

/

I

I

Chapter 23 •· Ray Optics

889

\ Formula and Magnification for Spherical Mirrors

we have, 1

Mirror fonnula 1

1

1

1

1

-=-+! v u where symbol possess their usual meanings.

1

1

1

Now,

I v f f-v m=-=-=--=-0 u f-u ·f where I = size of image perpendicular to principal axis 0 = size of object perpendicular to principal ruds.

Axial magnification

~~ = :: = ( f ~

J

2

-

=(f

r

~!

"

Areal magnification 2

1

1

1

-=----=-v l.Sf f 3f

Lateral magnification

max = -

1

-+-=-or--+-=-u v F -l.Sf v f

2

m = Ar =::.:_=(-!) =(f - v) ar Ao u2 f- u . f

ni

=-~"" __l_[_ =-2

u l.Sf h or __£ = -2 or h2 = -2h1 = -5 em. hl The image is 5 em long. The minus sign shows that it is inverted.

Instance 5 ·A concave mirror of focal length 10 em and a convex mirror of focal length 15 em are placed facing each other 40 em apart. A point object isplaced between the mirrors, on their common axis and 15 em from the concave mirror. Find the position of the image produced by the successive reflections, first at concave mirror and then at convex mirror. (a) 6 em (b) +10 em (c) + 15 em (d) +30 (.7n · Interpret According to given problem, for concave mirror. f= -10

M

where , A 1 = area of image A 0 = area of object.

Instance 4 An object of length 2.5 em is placed at l.5f from a concave mirror; where f is the focal length of the mirror. The length of the object is perpendicular to the principal a·ds. Find the length of the image. Is the image erect or inverted ? (a) -5 em (b) 5 em (c) 6 em (d, --6 em Interpret and

The focal length F u

10

=

-f

= -l.5f

1--f

• F

1 - - - -- 1.5 f ------;~

u=

~

15 em andf = - 10 em .!.+-- = -1- ie, v = -30em v -15 -10 ie, concave mirror will form real, inverted and enlarged image 11 of object 0 at a distanci! 30 em from it, ie, at a distance 40-30 = 10 em from convex mirror. For convex mirror the image 11 will act as an object and so for it u = -10 em andf = + 15 em. .!.+-1-=_.!._ ie,v= +6Cin v -10 15 So, final image 12 is formed at a distance 6 em behind the convex mirror and is virtual as shown in figure.

1

So,

lntext Que,~!lo~,. -2a.l M~'""

,

(i); Does the mirror fommla hold g?od for a plane mirrgr? .· , , · ·· . ..· _._. _ _. :,: , ..... (iit An object :is'placed between two ·plane parallel mirrors. Why do tpe t images ;~t fainter . ·· (iii)' Why are inir:rors used in search-lights parabolic and not concave spherical? , ; ·)> qy~ If yo~ were driving a car, what typ~ ofmirtoi would .you prefer to use (9_r;ob~~~il:affic .at ""''"''"•Adr?

w> Refraction When light passes from one medium, say air, to another medium, say glass, a part is reflected back into the first medium andthe restpa~,;, ·., .: .. ..;. ·n-''.': _,·: ;\;)assesintu the second medium, it either bends towards the normal or away from the normal. This phenomenon is known as refraction.

laws of Refraction (Snell's law) (i) If meditim 1 is a vacuum (or in practice air) we refer 1J.l2 as the absolute rE>frai'! h •" ir> ;2 V2

<

l-12

> l-11

ll :> < u l

"-2 < "-1

"-2 > "-1

VI

If 1-1-2 ::> 1-1-1 then v 1 > v 2 and A-1 > 1..2 , ie, in a rarer m~dium, speed and hence, wavelength of light is more . (iv) In general, speed of light in any medium is less than its speed in vacuum. It is convenient to define refractive index 11- of a medium as.

Interpret

Let the angle of incidence, angle of reflection and an,le of refraction be i , rand r', respectively. Now, as per the question 90° -r + 90°- r' = 90° => r' = 90°- i (because i = r) In case of reflection according to Snell's law, 1 sin i = 1-l sin r ' or sin i = 1-l sin (90° - i) => tan i = 1-l or i = tan-1 [!1-] = tan-1 (1.62) = 58 .3°

Instance 8 Refractive index of glass with respect to water is 1.125. If the absolute refractive index of glass is 1.5, find ,the absolute index of water. (a) 1.33 (b) 2.33 (c) 0.33 (d) 0.44 Interpret

Here, the refractive index of glass with respect to water

ie, wll-g = 1.125 and absolute refractive index of glass llg

= 1.5.

We know that

c v

Speed of light in vacuum 11

(b) 85 .3° (d) 65°

(c) 60°

= Speed of light in medium

Instance 6 Light is incident from air on oil at an angle of30°. After moving through oil-1, oil-2, and glass it enters water. If the refractive indices of glass and water are 1.5 and 1.3, respectively, find the angle which the ray makes with normal in water. Air Oil-1

Oil-2

Apparent Shift of an Object due to Refraction Due to bending of light at the interface of two different media, the image formation due to refraction creates an illusion of shifting of the object position. Consider an object 0 in medium. After refraction, the ray at the interface bends. The bent ray when it falls on our eyes, is perceived as corning from I. For nearly normal incident rays, 81 and 82 will be very small. Rarer

J 2 .6

(b) sin-

1 3.6

(d) sin- 1 (2 .6)

(a) sm -1 ( - 1 0

(c) sin-

Interpret

1

(-

J

1

J.lz

(_]__J 2.6

As we know 1-l sin i = (constant) !l-air sin ~air)

=>

=

1-l-glass sin r (glass) .

!l-air =--Sin lair 1-l-glass Again, ll·glass sin iglass = llwater sin r water From Eqs. (i) and (ii) sin 30 = 1.3 sin r 0

0

0

sm~glass)

1 2xl.3

1 2.6

0

... (i) ... (ii)

,

AB

.

AB

sm8 2 = - - - - - - - - - - - - - , - - - - - - Image distance from the refracting surface 1 2.6

smr=--=-, r=.sin-1 ( 0

·

tan 81 = sm 81 = -=-':-:-:-':__-:-:----:---.-=-:-----"':'--'---:----:-Object distance from the refracting surface Similarly,

J

sin 81 sin8 2

_ 11 2 :::} 11 - 1 2 - 11 1

_

AB AB _ 1 OB BI -

11 2 11 1

Chapter 23 • Ray Optics BI OB

-=

v~

Apparent depth J..lz ·= Real depth J..L1

So, Shift = Real depth- Apparent depth = Real depth ( 1 -

~~ J

Case I If ).ll < ).lz Shift becomes negative , image distance > object distance, ieJ image is farther from the refracting surface.

Total Internal Reflection

Whenever a ray of light goes from a denser medium to a rarer medium it bends away from the normal. As angle of incidence in denser medium increases, angle of refraction also increases in rarer medium. The angle of incidence in denser medium for whi.c h the angle of refraction in rarer medium is 90° is called the critical angle (C) . sin C = ~Lrarer sin 90° J..ldenser

Case II If ).ll > ).lz. Shift becomes positive, image distance < object distance, ie image is closer to the refracting surface. Case III If ).lz = 1 or ).! 1 ~

Shift = Real depth ( 1 -

~J

Instance 9 A fish in an aquarium, approaches the left wall at a rate of 3 ms-1, and observes a fly approaching it at 8 m.~- 1 . If the refractive index of water is (4/ 3), find the actual velodty of the fly.

~X ).lX

(a) 3.75 ms- 1 (c) 0.75 ms- 1

(b) 2.75ms-1 (d) 4.75 ms- 1

Interpret

For the fish, appa.rent distance of the fly from the wall of the aquarium is ).lX. If x is actual distance, then apparent

=~ J..ld

sinC=~

=>

J..ld

c = sin-1 (~:J

=>

).l

891

Now, if the angle of incidence in the rarer medium is greater than the critical angle (C), then the ray instead of suffering refraction is reflected back in the same (denser) medium. This phenomenon is called total internal reflection. For total internal reflection to take place following set of conditions must be obeyed. (i) The ray must travel from denser medium to rarer medium. (ii) The angle of incidence i must be greater than critical angle C.

Instance 11 An isotropic point source (bulb) is placed at a depth h below the water surface. A floating opaque disc is placed on the surface of water, so that the bulb is not visible from the surface. What is the minimum radius of the disc? Take refractive index ofwater=).l. Interpret

As shown in figure, light from bulb will not emerge out 1 of the water if at the edge of disc. ·

i>C

velocity will be d(J.!X) dt

sin i >sin C

... (i)

(v.pp)fly = !l vfly

Now, the fish observes the velocity of the fly to be 8 ms-1 • Therefore, apparent relative velocity = 8 ms- 1 1 Vfish + Cvapp)fly = 8 ms- ~ 3 + ).l vfly = 8 vf!y

= 5 x ~ = 3.75ms-1 4

Instance 10 A layer of oil 3 em thick is flowing on a layer of coloured water 5 em thick. Refractive index of coloured water is 5/3 and the apparent depth of the two liquids appears to be 36/7 em. What is the refractive index of oil? (a) 1.4 (b) 2.4 (c) 3 (d) 2 Interpret

Apparent depth (AD) = .EL + 2_ 5

J..lz

,

7 7

J..lz =- = 1.4 5

. c =-1 sm

So, Eq. (i) becomes R

1

--===>~R2 +hz J..l

.2_= 36 -3= 15

or

R

J..l

3

-=--+7 5/3 J..lz

or

. .

smt=-,=== Jiz +hz and

Ill ' 112

36

Now, if R is the radius of disc and h is the depth of bulb from it

7

or

h R>--

~J..l-1

892

Chapter 23 • Ray Optics

• Refraction from a Spherical Surface Spherical surfaces are of two types (i) Convex (ii) Concave

1

u'

Solving it we will get, v' = -35.4cm

~

1E'

v'

(4/ 3) = (3/ 2) v' ( -40)

==>

A' B'' (Jl v') . Now usmg - - =-1, A' B'

2 •()······;;,······················:I ················

0

llz_= llz

==>

=>

(Jl 2u')

A"B" = (3/ 2)(-35.4) => A"B" = 5.3cm (5 .3) (4/ 3)(-40)

The final images in all the above cases are shown in figure . I

2

·--·········-p·· ---------------------•--·-·······

For both surfaces refraction formula is given by ~-~= Jllz-1 v u R 11-1 2 is refractive index of second medium with respect to first. If 1-1 1 and 1-lz are refractive indices of first and second medium with respect to air, then, llz _ J-11 = 1-lz ·· 1-11 v u R

~'~

Lens

Lens is a transparent medium bounded by two curved surfaces. Lenses are of two types 1. Convex or convergent lens 2. Concave or divergent lens

1. Convex or Convergent Lens The traqsparent medium bounded by two bulging surfaces is called convex lens . It is ofthree types (as shown) .

instance 12 A linear object of length 4 em is placed at 30 em from the plane surface of hemispherical glass of radius 10 em. The hemispherical glass is surrounded by water. Find the final position and size of the image:

(a) Double-convex lens (b) 4.3 em (d) 2.3 em

(a) 5.3 em (c) 5 em

;nterpret

4

3

For 1" surface 1-11 = -,J-1 2 =- ,u =-20cm, 3 2

(b) Plano-convex lens

(c) Concavo-convex lens

2. Concave or Divergent Lens The transparent medium bounded by two hollow surfaces is called concave lens. It is of three types (as shown).

R = +lOcm ,

and 8"

5.3cm A"

(a) Double-concave (b) Plano-concave (c) Concavo-concave lens lens lens

1 - - - - - v'

----~

1-lz _ J-11 = (J.lz -1-11) v u R (3/ 2) - (4/ 3) = (3/ 2- 4/ 3) ==> v (-20) 10 ==> v = -30 em Using A' B' = ~ => A' B' = (4/ 3) (- 30) AB JlzU ( 4cm) (3/2) (-20) ==> A'B' = 5.3 em A'B' behaves as the object for plane surface 3 4 ' Jl 1 = ,Jlz=j and R=oo,u=-40 Using

2

Some Definitions Relating Lenses Optical centre The optical centre is a point within or outside the lens, at which incident rays refract without deviation in its path.

s

Chapter 23 • Ray Optics

8~~ l,i

Principal axis The straight line passing through the optical centre of lens is called principal axis of lens. 1

Principal focus

Lens has two principal foci.

(i) First principal focus It is a point on the principal axis of lens, the rays starting from which (convex lens) or appear to converge at which (concave lens) become parallel to principal axis after refraction.

Laws of Formation of Images by Lens

'

·' (i) The rays corning parallel to principal axis of lens pas~

through the focus after refraction. ·· · (ii) The rays corning from the focus ofltilS go parallel tot~ principal a~is of lens after refraction. ': (iii) The rays of light passing through optical centre go straight after refraction without changing their path.

Lens Maker's Formula ··.·..

.. ... _...-::· F1

If R1 and R2 are the radii of curvature of first and seconq refracting surfaces of a thin lens with optical centre C of foc~l length/ and refractive index 1 ~2 then according to Lens Maker'$ formula

(ii) Second principal ,focus It is the point on the principal axis at which the rays coming parallel to the principal axis converge (convex lens) or appear to diverge (concave lens) after refraction from the lens.

c u-----1...._--v

..!..=(1~2-1)(_..!__ _ _..!__] f

~ Both the foci of convex lens are real while that of concave lens are virtual.

R2

..!..=(!l-l)(_..!__ _ _..!__J f R1 R2

where, 1 ~ 2 = ~ is refractive index of material of lens with' respect to surrounding medium. . Thin lens formula is

1 1 1

-= - - ! v u

Focal length The distance between focus and optical centre of lens is called focal length of lens.

Table 23.3

Rl

Formation of Image by a Convex Lens

1.

At infinity

At the principal focus (F2) or Real, inverted and extremely in the focal plane diminished

2.

Beyond 2F1

Between F2 and 2F2

Real, inverted and diminished

894

Chapter 23 • Ray Optics

Real, inverted and of same size as the object

· 4.

Between F1 and

Beyond 2F2

Real, inverted and magnified

2Fl

5.

At F 1

At infinity

Real, inverted and highly magnified .

6.

Between F 1 and optical centre

On the same side as the object

Virtual, erect and magnified

Formation of Image by Concave Lens

Dividing Eq. (i) by (ii), we get

The image formed is always virtual, erect and diminished and lies between the lens and F2 for all positions of the · object.

Instance 13 The focal length of convex lens is 10 em in air. Find its focal length in water. (Given, llg. = 3/ 2 and llw = 4 / 3) (a) 10 em (b) 20 em (c) 30 em (d) 40 em Interpret

1 - -=CI!g -1)(2__2_J Rl

fair

and

1

(llg

ff'ater = llw -

1

Rz

XRl1 - Ri1 J

..

(i)

..• (ii)

fwater ( llg -1 "fair = llg f ll w -1

J

Substituting the values, (3 / 2-1) fwater = ( 312 )fair - - -1 4/ 3 = 4 fair = 4 X 10 = 40cm

Instance 14 An object is placed at a distance of 10 em to the left on the axis of a convex lens L1 of fo cal length 20 em. A second convex lens L2 offocal length 10 em is placed co-axially to the right of the lens L 1 at a distance of 5 em from it. Find the position of the final image and its magnification. (a) 163. em on the right of the second lens, 3.33 3 . (b) 163. em on the right of the second lens, 1.33 3 (c) 163. em on the right of the first lens, 1.33 3 (d) None of the above

Chapter 23 • Ray Optics Interpret

895

Important Features

Here, for 1" lens,

1

u1 = -10 em f 1 = 20 em 1 1

vl

u1

1 1. Power of lens P = - -f(inm)

---=-

!1

=>

vl 20 10 v1 =-20cm

=> L1

L2

~

o·

, o;:;;

· 0 2 '!

';.:

\i v ·1'2

\ '!

o

!

,1,.\! Scm 1

1

1

= 2:1_ Vz = 20 _29_ = _± = 1.33 u1 u2 10 3x25 3

Height of image = _!£__ = ~·: ' ' '' Height of object 00' u Substituting v and u with proper sign, ) ,; . . If -I v

,

u

L,P; n

2

The lateral, transverse or linear magnificauon produced by a lens is defined by Height of image I m= =Height of object 0 A real image II' of an object 00' formed by a convex lens is shown in figure.

Thus,

h

i=1

Magnification of Lens

or

i=l

Magnification of combination 50

00' 0 -u I v -=m=0 u v m=-

n

n

p =Pl +Pz + ... =

ie, final image is at a distance o; 163_ em on the right of the second lens. , 3 T)le magnification of the image is given by;

'.

L.-1

Power of combination

M = m1 xm2 x ... =11m

- + - = - => v2 =-=16-cm v2 25 10 3 3

m

~ala2 .

=

This is Newton's formula. 3 . If two or more lenses are placed in contact, then equivalent focal length of the combination.

1

For 2nd lens, - - - = Vz u2 fz Here, u2 = -(20 + S),J2 = 10cm 1

f

1 1 1 -=-+-+ .. ·= f !1 fz

ie, the image is virtual and hence lies on the same side of the object. This will behave as an object for the second lens. 1

100 j(incm) Power of convex lens is positive and of concave lens is negative . 2. If distance of an object from first focus of lens is a 1 and distance of image from second focus is a 2 , then its focal length. P=

=>

1 1 1 -=---

i=l

4. If two lenses offocallengthsf1 andf2 are separated by a distance x, then its equivalent focal length 1

1

1

X

-=-+ - - F !1 fz fdz Power of combination, P = P1 + Pz- x P1P2 Total magnification remains unchanged ie, m=m 1 xm 2 5. If a lens is made of a number of layers of different refractive indices, then number of images of an object formed by the lens is equal to number of different media. 6. Cutting of a lens (i) If a symmetrical convex lens of focal length f is cut into two parts along its optic axis, then focal length of each part (a plano convex lens) is 2f. However, if the two parts are joined as shown in figure, the focal length of combination is again f. 2(,

(a)

(b)

2f

f

(c)

f

(d)

I

896

Chapter 23 • Ray Optics (ii) If a symmetrical convex lens of focal length f is cut into two parts along the principal axis, then focal length of each part remains changed at f . If these two parts are joined with curved ends on one side focal length of the combination is f_ . But on joining two 2

parts in opposite sense the net focal length becomes (or net power = 0).

7. Silvering of a lens (i) Let a plano-convex lens is having a curved surface of radius of curvature R and has refractive index J.l· if its plane surface is silvered, it behaves as a concave mirror of focal length.

~

f

R =- 2(~-t- 1)

(ii) If the curved surface of plano-convex lens is silvered

then it behaves as a concave mirror of focal length.

f

R =- 2!l

(iii) If one surface of a symmetrical double convex lens (R1 = R2 = R) is silvered, then the lens behaves as a concave mirror of focal length

f= (a)

Table 23.4

8 . The tabular difference between lens and mirror is given (b)

(c)

(d)

in table.

Difference between Lens and Mirror

1.

Convex lens

+ ve

+ ve

converging

2.

Concave mirror

- ve

+ ve

converging

3.

Concave lens

- ve

-ve

diverging

4.

Convex mirror

+ ve

-ve

diverging

,

R

2(2J.l-l)

897

Chapter 23 • Ray Optics Instance 15 A convergent lens of 6 D is combined with a diverging lens of -2 D. Find the power and focal length of the combination. (a) 26 em (b) 20 em (d) 25 em (c) 30 em Interpret Here, P 1 = 6 D, P2 = -2 D Using the formula, P = P 1 + P2 = 6- 2

Instance 16 A convex lens of 10 em focollength is combined with a concave lens of 6 em focollength. Find the focollength of the combination. (a) -15 em (b) 15 em (c) 10 em (d) -10 em Interpret Here,f1 = 10 cm,f2

.!. = ..!._+..!._ = _.!__.!_ = _ _.!_

Use the formula

= 4D

= -6 em, F =?

!1 ! 2

' F

f = liP= 1/4 m = 25 em.

10

6

llR and consequently Ov > Ow

898

Chapter 23 • Ray Optics Instance 18 Find the dispersion produced by a thin prism of 18° having refracting index for red light = 1.56 and refractive index for violet light = 1.68. (a) 2.16° (b) 1.16°(c) 3.16° (d) 2.10° Interpret

We know that dispersion produced by a thin prism

e = C!-lv -11R)A Here, 1-lv

Angular Dispersion It is the angular separation between the fwo extreme rays. Angular dispersion 8 = 8v - 8R = C11v - 11R )A Dispersive Power The dispersive power of a prism material is measured by the ratio of angular dispersion to the mean deviation suffered by light beam. :. Dispersive power ro- Ov -oR _ 1-lv -1-lR

--0--~,

where 11 is the mean value of refractive index of prism. The dispersive power of a prism depends only on its material and is independent of angle of prism, angle of incidence or size of the prism. Dispersive power is a unitless and dimensionless term. Dispersive power of a flint glass prism is more than that of a crown glass.

Dispersion without Deviation (Direct Vision Prism) 1. To produce dispersion witho~t mean deviation we use a combination of two prisms of different materials such that

= 1.68,~-tR = 1.56 and A

= 18°

e.=(1.68 -1.56) x 18° = 2.16° Instance 19

Calculate the dispersive power for crown glass from

the given data 1-lv

= 1.523

(a) 0.01639 (c) 0.05639

and 1-LR = 1.5145 (b) 1.05639 (d) 2.05639

Interpret

Here, 1-lv = 1.523 and 1-lR = 1.14S. . . d l.S23+1.S145 Mean re fr actlve m ex, 11 = l.S187S 2

Dispersive power ro is given by, ro = !lv -llR = l.S23 -l.S14S (!l-1) (l.S187S-1)

_ 0 01639

Instance 20 A prism of crown glass with refracting angle of so and mean refractive index = 1.51 is combined with a fiint glass prism of refractive index = 1.6S to produce no deviation. Find the angle of fiint glass . (a) 3.92° (b) 4.68° (c) 5.32° (d) 7.28° Interpret

Let A' be the angle of flint glass prism.

Here, A

= so and J-l = 1.51 for crown glass prism.

8 = (!l-1)A= (l.S1-1)xS = 2.SS 0

Deviation produced by flint glass

o' = (!!' - 1)A' = (1.6S -1)A' = 0.6SA'

For no deviation, 8' = 8 or 0.65A'

=

2.55

A'= 2.S5 = 3.92o 0.65

> Optical Instruments

A'= -(l-l.-1JA !-l-1

2. Net dispersion caused = C11v- 11R) A + C11' v - 11' R)A' = (11- 1)A (m - m') = 8 (m - m')

Deviation without Dispersion (Achromatic Prism) 1. To produce deviation

without dispersion we use a combination of two prisms of different materiaPs such that ~A'=- [!-lv -1-lR] ·A [!-l~ -!-l~]

2. Resultant

~eviation produced =·O[1-:]

Optical instrument is a device which is made from proper combination of mirrors, prisms and lenses. The principle of working of optical instruments depends on laws of reflection and refraction of light.

Microscope It is an optical instrument which forms a magnified image of a small nearby object and thus, increases the visual angle subtended by the image at the eye so that the object is seen to be bigger and distinct. )

..

l

~

(i) Simple microscope A simple microscope is a convex lens of short focal length which is fixed in a frame provided with ', handle . . ·c··j

Chapter 23 • Ray Optics Magnification of simple microscope (a) When final image is formed at least distance of distinct vision,

899

Object

D

M=1+-

f

(b) For relaxed eye, M = D

f where D

=

least distance of distinct vision.

Magnification of astronomical telescope

Figure shows a simplified version of a compound microscope . It consists of two converging lenses arranged coaxially. The one facing the object is called objective and the one close to eye is called eye piece. The objective has a smaller aperture and smaller focal length than those of the eye piece. (ii) Compound microscope

(a) For relaxed eye, M = _ fo ~ fe In this position, length of telescope L~ =fo +fe

(b) When final image is formed at least distance of distinct vision

J

M = _ f o ( 1 + fe D fe D Length of telescope LD =fo+ue f o = focal length of objective lens f e = focal length of eye piece

h

Magnification of compound microscope (a) For relaxed eye

M~ = _ vo (E-J Uo

fe

In this position, length of microscope L~ = Vo

(ii) Terrestrial telescope In an astronomical telescope , the final image is inverted with respect to the object. To remove this difficulty, a convex lens of focallengthf is included between the objective and the eye-piece in such a way that the focal plane of the objective is a distance 2f away from this lens.

+ fe

(b) When final image is formed at least distance of distinct

B

1~~~~~ A

VlSlOn.

MD

=~(1+ DJ fe Uo

Length of microscope, L0 =v0 +Ue v0 = distance of first image from object lens. U 0 = distance of object from objective lens. f. = focal length of eye piece.

Telescope Telescope is an optical instrument which increases, the visual angle at the eye by forming the image of a distant object at the least distance of distinct vision, so that the object is seen distinct and bigger. (i) Astronomical telescope It consists of two converging lenses placed coaxially. The one facing the distant object is called the objective and has a large aperture and large focal length. The other is called the eye-piece, as the eye is placed closed to it. The eye-piece tube can slide within the objective tube, so that the separation between the objective and the eye-piece may be varied.

'

~-r0 ~2r--- e r-~

Magnification of terrestrial telescope (a) For relaxed eye,

M~ =

1

In this position, length of telescope L~ = f o + 4f + f e (b) When final image is formed at least distance of distinct vision,

J

M = fo (1+ fe D fe D. Length of telescope , n

LD = fo + 4 f + de f o = focal length of objective lens f e = focal length of eye peice Galilean telescope A simple model of Galilean telescope is shown in figure. A convergent lens is used as the objective and a divergent lens as the eye-piece.

900

Chapter 23

* Ray Optics

~--------

fo _ _ _ _ _ ____..,

8

Resolving Power of a Microscope Resolving p;Jwer of e. microscope is defined as the reciprocal of the least separation between two close objects, so that they appear just separated, when seen through the rnicroscope. The least separation between two objects, so that they appear just separated is given by

1 A

d=--"'2J.1sin8 where fl is che refractive index of the medium between the objective of the microscope and the object. This distance is called limit of resoh~tion of the microscope.

Magnification of Galilean telescope (a) For relaxed eye,

M~ =

fo

fe

In this position, length of telescope L, = fo- fe (b) When final image is formed at least distance of distinct vision M = fa ' D fe

Resolving power of a micro5mpe = .!_ = 2J.1 sine d lc 8 = half angle of the cone of light from the point object, fl sin 8 = numerical aperture

(1 - fe J D

....de._____ _____ _

Length of telescope

0

Lv = fo -u,

Instance 21 An object is seen through a simple microscope offocal length 20 em. Find the angular magnification produced if the image is formed at 30 em from the lens. (a) 2.08 (b) 2.05 (c) 3.08

Interpret and

(d) 1.5 1.

Resolving power of microscope increases with increase in the value of the refractive index of the meaium between objective and object that's why oil immersion objective microscopes are used to achieve high resolving power. 2. The resolving power of microscope increases, with decrease in the value of the wavelength of the light used to illuminate the object, so microscopes using ult~aviolet light for illu(Tlinating the objects are used to achieve high resolving power. These are called ultra microscopes. Higher resolving power is obtained in electron microscope.

Given,f = + 20 em v = - 30 em ,

Using thefonnula, 1 -30

.!._.!.=.!. v

u

f

wehave,

1

1 20 u0 = 12 em

-U 0

'{angular magnification, M

=.!!_ = 25 =2.08 uo

12

Instance 22 A galilean telescope is 27 em long when focussed to form an image at infinity. If the objective has a focal length of 30 em, what is the focal length of the eye piece? (a) 3 em (b) -3 em (c) 2 em (d) -2 em Given,fo = + 30 em. Objective Eye-piece Length of telescope is given 27cm. Therefore, u, = + 3 em. For the final image at infinity, the intermediate image should lie at first focus of eye piece of the I-- 27 em --l~ Galilean telescope. 1---~- 30 em •I fe=-3cm

Interpret

> Resolving Power of Telescope Resolving power of telescope is defined as the reciprocal of the smallest angular separation between two distant objects, so that they appear just separated, when seen through the telescope. The smallest angular separation between two objects, so that they appear just separated is found to be de= 1. 2 21c D

where D is the diameter of objective Resolving power of telescope

D

1.22/c

lntext Questions 23.3 en does a ray fucid~nf on _a prism deviate away from the base? "ngs) observed sometimes round the sun or moon? f gla.~s for lights ?f yellow, green and red colours are f!y. llg and flr respectively. Rearrange 'thes~ fv~l~7 1 : _; e position a object relative to a biconvex lens so that it behaves like magnifying lens? erted, ~ill it serve as a microscope?

of

901

Chapter 23 • Ray Optics

Chapter Compendium 10. Refraction at Sphe,rical Surfaces,

are o\:1eyed at every reflectin,g surface.

(i) For a spheric1il surface, f.Lz. ..., f.LI =

Hz - f.L1

ofsy:mmetrical spherical surfaces, are two (i) convex, and (ii) concave. which the reflection takes place at the bulged a convex mirror and the mirror in which place at the depressed surface is called a

(ii) Magnification, m

= 111::_ flzU

(iii) When the object is

relation can be obtainted relation becomes

vv Ill Ill 812 > 813 Smaller the value of critical angle more the chance of total internal reflection.

I

34. Using, llz - 111 = llz -Ill , we get v u R 1.5 1.0 1.5 -1.0 20 v v=±60cm or ' 1 1. 1 35. As !l2':-->-->~2':J2=1.414. sine sin45° 11 -v2 :. Possible value of J..l are 1.5 and 1.6.

,

= ( 1.5 - 1) X 4 =4.5-4 --x4=0.5D. 4 413

A+B 1 37. !l=~ => ll""~

:. __.!__=(1.5-l)(I__I_) 20 = R 1

·

In water

4000A will greater

(~ - 1 )[__.!__ - __.!__] !lm Rl Rz

Since (IlL< !lm) because lens is of water and IlL= llw · 39. !l=-1-= 1 _1__ _± sine sin48 .6 0.75 3 40. Light cannot undergoes total internal reflection when it is travelling from air to water, ie, from rarer to denser medium. · 1 41. sine=_.!_= - - = = 0.6667 ll 3/2 3 e = sin- 1(0.6667) = 41.8° 1 - ,s1n · e=l 42 . From ll = - . Sill e !l As llv > !lr ev< e,. 43. As shown in figure OS= h When angle of incidence is slightly greater than --------e, light undergoes total ------------------· internal reflection. ~-=-=~-=-=-h c --------~~-=~-=-=· -------:. Diameter of circle of -----------------------------· light coming from water ----------------surfaee = 2 r = 2 (OB) ------- ~ --·s----------· = 2 OS tan e = 2h tan C. 44. In air or water, a convex lens made of glass behaves as a convergent lens but when it is placed in carbon disulfide, it behaves as a divergent lens. Therefore, when a convergent lens is placed inside -a transparent medium of refractive index greater than that of material of lens, it behaves as a divergent lens. It simply concludes that property of a lens whether the ray is diverging or converging depends on the surrounding medium. 45. If a mirror is placed in a medium other than air its focal length does not change asf = R/2, but for 1,the lens

3.

,------~---_-,

_.!._ = Cang -1)(__.!__ _ __.!__) fa Rl Rz · __.!__ = C.vng -1)(__.!__ _ __.!__) fw Rl Rz As w ng lla·

and

932

Chapter 23 • Ray Optics

46. The magnifying power of telescope in relaxed state is . , m = fo ' h .'.s So, for high magnification, the focal length of objective length should be larger than of eye-piece.

53. When lens is in air, then ..!_=(aJ.l. -1)(_!_ _J_J.· f g ' R R2 · ' l

)

Resolving power of a telescope· =

or

_d_:..

1.22A For high resolving power, diameter (d) of objective should be higher. 47. We know that power of lens is a reciprocal of its focal " 1 1 . = 2D length, hence P = - = 50 f-

1 ' ) (1 1J ~=(1.5-1 --10 R1 R2 .

(;1- ;J=~

or

When lens is in water, then

~f = r:Jlgllw -1J(_!_ _ _!_J Rl Rz

100

Since, lens is concave hence, its power will be 2D. If the objective is placed at infinity then u==,v=?,f=50cm 1 1 1 From the formula, - - - = v 1

I= fo a

1

-50 v=-SOcm Thus, concave lens will form an image of the object at infinity at a distance of 50 em. 48. When glass surface is made rough, then light incident on it is scattered in different directions. Due to which its transparency decreases. There is no effect of roughness on absorption of light.

.J6 1 .J3 =sine " 1 " so smC=-=sm4

fe

-~- = _Q2_ => ~ = so

00

149. Refraction index of diamond w.r.t. liquid zlld = -_, smC

or

f'=40cm 54. For a telescope

f

u 1

v

=(~-1 )x!=_..!._ 1.33 5 40

0.5°

0.03

55. !tis observed if Li=Le=~LA 4 . _A+om and [=--- . 2 Here, LA= 60° 3 . and L t=Le=-LA 4

om = 2x~x60°-60° = 30° 4 56. Incident ray and finally reflected ray are parallel to each other means, o= 180°

F2

c = 45° 50 Red glass transmits only red light and absorbs all the colours of white light. Thus, when green flower is seen through red glass it absorbs the green colour, so it appears to be dark. 51. Here, u = -mf, u = v,f = -f Using lens formula !_..!. = ..!., we have v u f or

1 1 ( 1+1) =-(m+1)=-(m+1) -1 1 -=-v f m mf u

or

-=--

v u

From => => 57.

1

Jl

f

X

-X

X

:. Coordinate of point Pare (2f, 2f) . f

1

oc-

m+1

. . =v = -1Lmear magm'fi catiOn u (m+ 1) 52. The lens formula is 1 1 1 -=--... (i) ! v u The graph between u and vwill be curve as shown in figure . Let the coordinates of P be (x, x), then u = -x and v = x 1 1 1 2 - = - - - = - orx = 2f FromEq.(i),

0=360°-20 180° = 360° -20 e = 90°

A.

1-lrarer

<

1-ldenser

< Ararer ie, wavelength decreases. Actenser

58.

allg

1 1 =-.- -=>sinC=--. smC allg

As J..l for violet colour is maximum, so sin C is minimum and hence critical angle C is minimum for violet colour. 59. As A.r > lvv :. ov > or

The statement given is true neither, in prism spectrum nor in grating spectrum.

~ I

II

Chapter 23 • Ray Optics

. ·. 60. _!_ = F

-j

t

'

i

C~-t-1) (__!_ - __!_ J· ·R R

."',.

0.06

R1

R2

100 3

f

R 1 Now, -1= - or R 2 = 2R 1 R2 2

100

l . 71. 8=-,8=1 mm.

3

r

R1 = = 0.045m 200 R 2 = 2R 1 = 2 x 0.045 m = 0.09 m.

So,

and

1 - = ~ = 0.75 4/ 3 4

61. sinC = _!_ = -

c

3

70. The angle subtended at the eye becomes 10 times larger. This happens only when the tree appears 10 times nearer.

9

11

I

20 -20 1 1 1 =-+-=40 120 30 4 = -x30 = 40cm

(Taking R1 positive and R2 negative) 1 1 1 -+-=---R1 R2 0.06 X 0.5

. Rz I

!

=(~-1)_(-~-~)

1 - - · = (1.5 -l)(_!_+_!_J

3 2R1

'

R1

2

1

933

1 1t 1 8 = - = - x - rad 60° 180 60 1=3m l x=r=8 3

= 48°

1 = 10 krn.

1t

:. Angle with horizontal = 90°- C = 90° - 48° = 42° 62. AI> it is clear from figure distance travelled in. glass place

-

180 60 72. For dispersion without deviation

81 - 82 = 0, ie , 81 = o2

=DB= OC

cosr

x-

or

Cl-11 -l)Al = C~-tz -l)A2 (1.54 -1)4° = (1.72-1)A2

OB=-t_ .

cosr 90° 2

63. AI> the ray turns through 90°, therefore, C = - = 45°

~-t=-1-=_1_= sin C

sin 45°

lin =J2=1.414. 1/ -v 2

64. p = _!_ = (!l-1)(_!_ _ _!_J F R1 R2

= (1.4 -1)(_!_ + 100) = 40 = 2.66D 15 15 00

73. The image formed by a concave mirror is always real when the object is placed between focal point and infinity. But when the object is placed between focal point and pole, then the image is always virtual. But when the object is virtual ie, object is placed far away from the focal point, then the image is certainly real. Hence, the required answer is certainly real if the object is virtual. 74. Given that, the refractive index of the lens w.r.t. air, aflg = 1.60 and the refractive index of water w.r.t. air aflg = 1.33 The focal length of the lens in air, f = 20 em. We know that for a lens

_!_ =

f

C~-t-1) (__!___!_ J R1 R2

When the lens is in the air

66.

~-t=

__!_ = (

sin(A+om)/2 sin(60°+60°)/2 = sinA/2 sin60° /2

= sin60° = .J3 = 1. 73 sin30° .2xl / 2 6 7. From the lens formula

!.. -!.. = _!_ = constant

v u f u is always negative, vis positive. 68. The distance in the experiment are measured by a vernier scare provided on the microscope .

20

a

~l

g

-1)(_!_- __!_ R1

1

R2 )

or

-=(1.60-1) 1 (1 -1 J 20 R1 R2

or

__!_ = 0.6ox(_!_ _ _!_J

20 R1 __ R2 When the lens is in the wate~

__!_=( 11 -l)(_!_ _ _!_J f' w g R1 R2

... (i)

934

Chapter 23 • Ray Optics

;.-(:~: 1)(~1 ~2 )

or

-\ = ( a !-lg -a 1-lw f a!-lw

or

)(J__J_) R1

81. Bi-convex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens. Focal length of bi-convex lens ]_ = (n

f

Rz

]__ = (1.60 -1.33 )(__!_ ___!_) f' 1.33 R1 R2 ]__- 27 (__!_ ___!_ ) · ...(ii) f' 133 R1 R2 On dividing Eq. (i),by Eq. (ii), we get .[_ 0 .60x133 20 ·. 27 or f' = 20x2 .95cm ~ 60cm. Hence, its focal length is three times longer than in air. d 25 75. M=1+-=1+-=6

f

5

77. om= A,!J- = 1.5,A =? sin(A + om)12 sin(A + A)l2 1-1 = sinAI2 = sinA12 2 sin A I 2 cos A I 2 sinAI2

3 cos-=-

f=-R-.

1

(%}A= 2cos-

1

(%)

For plano-convex lens ]__ = (n

f1

=x(1-~J=6(1- 3 ~ 2 )

R

R

!1 =

Comparing Eqs. (i) and (ii), we see that focal length becomes double. As power of lens P ex:

1 -,--......,-,---:-

focal length

Hence, power will become half. New power =

.±2 = 2D

82. In normal adjustment, L = fo + fe = 200 + 4 = 204 em.

1-lv > 1-lr

ov >or or D 2 > D1 or D1 > D 2 84. As object is at the centre of the sphere, the image must be .

. 1 1 8 60 85. From smC=-=--=0.75,C=4. 1-1 1.33 d

1

=6 x -=2 em 3

Magnification m = fo = 4 fe

fo = 4fe I

5fe-10 em fe = 2 em f a= 8 em fa = 8 em, fe = 2 em Hence, L4 and L 1 will be used.

,

... (ii)

(n-1) ·

:. Distance of virtual image from centre of sphere = 6 em.

80. Length of tube 10 em f a + fe = 10 em

or and

.!..) =

-1)(]_-

~~~~00~

3 2 =9xi 2 cm J2 = 1

Putting in Eq. (i),

... (i)

2(n -1)

ie

78. In displacement method, size of object, 0 = ~1 1 12

79. Shift

R

4

~ = cos..

f

Rz

As 1-lv = 1.525 and 1-lr = 1.520

2

A 2

2

83. For o = (!J-- 1)A

3 A -= 2cos -

2

R1

1

-=(n-1)-

·or

76. The limit of resolution of an optical instrument arises on , account of diffraction of light.

-1)(2_- __!__)

= 2 x tan C = 2

x 8 tan 48.6°

= 16 x 1.13 = 18.08 em

1inm -4 86. A=--=5x10 mm 2000 = 5 x 10-7 m = 5000 A 5000 A.'= 2:. = = 40ooA 1-1 1.25 87. When a ray of light moves from one medium to other, its velocity changes. This change depends on refractive index of the medium. Light travels from denser to rater medium, ie, from medium of higher refractive index to lower refractive index. So, in second (rarer) medium, its velocity in.creases. ~

.